Search the corpus

    A variety of positive dimension has the cardinality of $k$

    P-AGH48CARDINALITY

    Area
    Algebraic Geometry
    Topics
    Birational Geometry, Varieties, Zariski Topology
    Status
    draft

    1. Show that any variety of positive dimension over \(k\) has the same cardinality as \(k\).

    2. Deduce that any two curves over \(k\) are homeomorphic.

    Hint

    Treat \(\AA^n\) and \(\PP^n\) first. Then for any \(X\), use induction on the dimension \(n\): make \(X\) birational to a hypersurface \(H \subseteq \PP^{n+1}\), and show that the projection of \(H\) to \(\PP^n\) from a point not on \(H\) is finite-to-one and surjective.

    Solution

    1. For every integer \(n\ge1\),\[ \abs{\AA^n}=\abs{\PP^n}=\abs{k}. \]

    The algebraically closed field \(k\) is infinite. For every finite \(n\ge1\), finite Cartesian powers of an infinite set have the same cardinality as the set itself, so \[ \abs{\AA^n}=\abs{k^n}=\abs{k}. \]

    Projective space is the finite disjoint union of its standard affine strata \[ \PP^n = \AA^n\sqcup\AA^{n-1}\sqcup\cdots\sqcup\AA^0. \] Thus \[ \abs{\PP^n}=\abs{k} \] as well.

    2. Let \(X\) be a variety of dimension \(d\ge1\). Then\[ \abs{X}\le\abs{k}. \]

    By definition, \(X\) is quasi-projective, so it is a locally closed subset of some projective space \(\PP^N\). Hence \[ \abs{X}\le\abs{\PP^N}=\abs{k} \] by step 1.

    3. The same variety satisfies\[ \abs{X}\ge\abs{k}. \]

    Choose a nonempty affine open subset \(U\subseteq X\). Because \(X\) is irreducible, every nonempty open subset has the same function field and hence the same dimension, so \[ \dim U=d. \]

    By the affine Noether-normalization theorem in T-MORFIBDIM, there is a finite morphism \[ \pi:U\to\AA^d \] induced by an inclusion \[ k[t_1,\ldots,t_d]\hookrightarrow A(U). \] The inclusion makes \(\pi\) dominant. A finite morphism is closed, so its image is both dense and closed in \(\AA^d\); hence \[ \pi(U)=\AA^d. \] Thus there is a surjection of underlying sets \[ U\twoheadrightarrow\AA^d. \] Therefore \[ \abs{X}\ge\abs{U}\ge\abs{\AA^d}=\abs{k}. \]

    4. Every positive-dimensional variety has cardinality\[ \boxed{\abs{X}=\abs{k}}. \]

    3.This proves (a).

    5. Every curve over \(k\) has the cofinite topology on its set of closed points.

    Let \(C\) be a curve and let \(Z\subsetneq C\) be closed. Write \(Z\) as the finite union of its irreducible components; this is possible because varieties are Noetherian. Every irreducible component of \(Z\) is a proper irreducible closed subset of the one-dimensional irreducible variety \(C\), so it has dimension zero. An irreducible zero-dimensional variety is a single point. Thus every proper closed subset of \(C\) is finite.

    Conversely, every point is closed over the algebraically closed field \(k\), so every finite subset is closed. Hence the closed subsets of \(C\) are exactly \(C\) and the finite subsets.

    6. Any two curves over \(k\) are homeomorphic.

    Let \(C\) and \(D\) be curves. By part (a), \[ \abs{C}=\abs{k}=\abs{D}, \] so choose a bijection \[ b:C\to D. \] By step 5, both spaces carry the cofinite topology. A bijection carries finite sets to finite sets, so both \(b\) and \(b^{-1}\) preserve closed sets. Therefore \(b\) is a homeomorphism. This proves (b).

    7. Q.E.D.

    Steps <1>1–<1>4 prove (a), and steps <1>5–<1>6 prove (b).

    Source Collections

    Dependencies

    • T-MORFIBDIMFibre dimension and upper semicontinuityuses